【POJ 2348 Euclid's Game】 + 博弈

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本文介绍了一个基于数学原理的两人游戏——欧几里得游戏，并提供了详细的算法实现过程。游戏由两名玩家交替进行，目标是通过减法操作使数字变为零。文章通过示例解释了游戏规则，并给出了一个判断胜负的算法。

Euclid’s Game
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9172 Accepted: 3756

Description
Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7):

an Stan wins.

Input
The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.

Output
For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed.

Sample Input

34 12
15 24
0 0

Sample Output

Stan wins
Ollie w

若出现 N - M >　Ｍ　情况是否为自由态？

ＡＣ代码：

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
int main()
{
    LL N,M;
    while(scanf("%lld %lld",&N,&M) != EOF){
        if(N == 0 && M == 0) break;
        bool ok = true;
        while(1){
            if(N < M) swap(N,M);
            if(N % M == 0 || (N > 2 * M)) break;
            N -= M;
            ok = !ok;
        }
        if(ok) printf("Stan wins\n");
        else printf("Ollie wins\n");
    }
    return 0;
}